直角三角形ABE中tanB=AE/BE,因E为BC中点,tanB=AE/(BC/2)=2AE/BC=2,因此AE=BC,由于平行四边形对边相等AD=BC,因此AE=AD.
2 ∵在□ABCD中,AD∥BC,AE⊥BC于E ∴AE⊥AD于A,∠FPE=∠ADP ∵AD=AE,∠EAD=90° ∴将△AEF绕点A逆时针旋转90°得到△ADG ∴△AEF≌△ADG,∠FAG=90° ∴AG=AF,∠ADG=∠AEF ∵EF⊥PD,AE⊥BC ∴∠AEF+∠PEF=90°,∠FPE+∠PEF=90° ∴∠AEF=∠FPE ∵∠ADG=∠AEF,∠FPE=∠ADP ∴∠ADG=∠ADP ∴点G在PD上 ∵AF=AG,∠FAG=90° ∴FG=根号2AF ∵FG=DF-DG=DF-EF ∴DF-EF=根号2AF
3 DF+EF=根号2AF
DF-EF=根号2AF